Let

\[f''++2g=4\]

(1)\[f+g=0\]

(2)We can obtain a single equation in

\[f\]

by differentiating in (2) and substituting into (1).\[f''++2g=4\]

(1)\[f'+g'=0\]

(3)From (3)

\[f'=-g'\]

then we can write (1) as \[f''-2f'=4\]

This last equation can bee solved using the integrating factor method.

Multiply the equation throughout by

\[e^{\int{-2dx}}= e^{-2x}\]

.We obtain

\[f' e^{-2x}-2f e^{-2x}=4e^{-2x} \]

.We can rewrite the equation as

\[(fe^{-2x})'=4e^{-2x}\]

.Integrate both sides.

\[fe^{-2x}=-2e^{{-2x}}+ C\rightarrow f=-2+Ce^{2x}\]

then from (2) \[g=2-Ce^{2x}\]

.